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Question: If \[f\left( \theta \right) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}\] then th...

If f(θ)=1−sin⁡2θ+cos⁡2θ2cos⁡2θf\left( \theta \right) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }} then the value of f(11∘)⋅f(34∘)f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right)
A) 12\dfrac{1}{2}
B) 34\dfrac{3}{4}
C) 14\dfrac{1}{4}
D) 1

Explanation

Solution

We will solve this question by splitting the numerator in two parts and then applying various trigonometric formulas and half angle formulas of sin⁡θ\sin \theta and cos⁡θ\cos \theta respectively.

Formulas used:
We will use following formulas:

  1. sin⁡2θ=2sin⁡θ⋅cos⁡θ\sin 2\theta = 2\sin \theta \cdot \cos \theta
  2. cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
  3. sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
  4. (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
  5. (a2−b2)=(a−b)(a+b)\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)
  6. tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
  7. tan⁡(a−b)=tan⁡a−tan⁡b1+tan⁡a⋅tan⁡b\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a \cdot \tan b}}

Complete step by step solution:
According to the question,
f(θ)=1−sin⁡2θ+cos⁡2θ2cos⁡2θf\left( \theta \right) = \dfrac{{1 - \sin 2\theta + \cos 2\theta }}{{2\cos 2\theta }}
Now, splitting the numerator of R.H.S. in two parts, we get
⇒f(θ)=1−sin⁡2θ2cos⁡2θ+cos⁡2θ2cos⁡2θ\Rightarrow f\left( \theta \right) = \dfrac{{1 - \sin 2\theta }}{{2\cos 2\theta }} + \dfrac{{\cos 2\theta }}{{2\cos 2\theta }}
We will now use the trigonometric formula sin⁡2θ=2sin⁡θ⋅cos⁡θ\sin 2\theta = 2\sin \theta \cdot \cos \theta and cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta to expand the terms of the above equation.
Therefore, we get
⇒f(θ)=sin⁡2θ+cos⁡2θ−2sin⁡θ⋅cos⁡θ2(cos⁡2θ−sin⁡2θ)+cos⁡2θ2cos⁡2θ\Rightarrow f\left( \theta \right) = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta - 2\sin \theta \cdot \cos \theta }}{{2\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)}} + \dfrac{{\cos 2\theta }}{{2\cos 2\theta }}
As we can notice, the numerator is in the form of a2+b2−2ab{a^2} + {b^2} - 2aband the denominator is in the form of (a2−b2)\left( {{a^2} - {b^2}} \right), hence, applying the formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab and (a2−b2)=(a−b)(a+b)\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)in the numerator and denominator respectively, we get
⇒f(θ)=(cos⁡θ−sin⁡θ)22(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)+cos⁡2θ2cos⁡2θ\Rightarrow f\left( \theta \right) = \dfrac{{{{\left( {\cos \theta - \sin \theta } \right)}^2}}}{{2\left( {\cos \theta - \sin \theta } \right)\left( {\cos \theta + \sin \theta } \right)}} + \dfrac{{\cos 2\theta }}{{2\cos 2\theta }}
Now, eliminating the same parts of numerator and denominator in RHS, we get
⇒f(θ)=(cos⁡θ−sin⁡θ)2(cos⁡θ+sin⁡θ)+12\Rightarrow f\left( \theta \right) = \dfrac{{\left( {\cos \theta - \sin \theta } \right)}}{{2\left( {\cos \theta + \sin \theta } \right)}} + \dfrac{1}{2}
Taking 12\dfrac{1}{2} common, we get
⇒f(θ)=12[(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)+1]\Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {\cos \theta - \sin \theta } \right)}}{{\left( {\cos \theta + \sin \theta } \right)}} + 1} \right]
Dividing numerator and denominator by cos⁡θ\cos \theta , we get
⇒f(θ)=12[(cos⁡θcos⁡θ−sin⁡θcos⁡θ)(cos⁡θcos⁡θ+sin⁡θcos⁡θ)+1]\Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {\dfrac{{\cos \theta }}{{\cos \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}}{{\left( {\dfrac{{\cos \theta }}{{\cos \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }}} \right)}} + 1} \right]
Now, using tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} in the above equation, we get
⇒f(θ)=12[(1−tan⁡θ)(1+tan⁡θ)+1]\Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {1 - \tan \theta } \right)}}{{\left( {1 + \tan \theta } \right)}} + 1} \right]
Substituting 1=tan⁡45∘1 = \tan 45^\circ in the above equation , we get
f(θ)=12[(tan⁡45∘−tan⁡θ)(1+tan⁡45∘⋅tan⁡θ)+1]f\left( \theta \right) = \dfrac{1}{2}\left[ {\dfrac{{\left( {\tan 45^\circ - \tan \theta } \right)}}{{\left( {1 + \tan 45^\circ \cdot \tan \theta } \right)}} + 1} \right]
Using the formula tan⁡(a−b)=tan⁡a−tan⁡b1+tan⁡a⋅tan⁡b\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a \cdot \tan b}}, we can simplify above equation as

⇒f(θ)=12[tan⁡(45∘−θ)+1] \Rightarrow f\left( \theta \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - \theta } \right) + 1} \right]……………………………(1)\left( 1 \right)
Now we will find f(11∘)⋅f(34∘)f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) by substituting the value of θ\theta as 11∘11^\circ and 34∘34^\circ the equation(1)\left( 1 \right).
f(11∘)=12[tan⁡(45∘−11∘)+1]f\left( {11^\circ } \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 11^\circ } \right) + 1} \right]
and
f(34∘)=12[tan⁡(45∘−34∘)+1]f\left( {34^\circ } \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 34^\circ } \right) + 1} \right]
Multiplying the two equations, we get
f(11∘)⋅f(34∘)=12[tan⁡(45∘−11∘)+1]⋅12[tan⁡(45∘−34∘)+1]f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 11^\circ } \right) + 1} \right] \cdot \dfrac{1}{2}\left[ {\tan \left( {45^\circ - 34^\circ } \right) + 1} \right]
⇒f(11∘)⋅f(34∘)=14[tan⁡(34∘)+1]⋅[tan⁡(11∘)+1]\Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {\tan \left( {34^\circ } \right) + 1} \right] \cdot \left[ {\tan \left( {11^\circ } \right) + 1} \right]
Now, multiplying both the brackets on RHS, we get
⇒f(11∘)⋅f(34∘)=14[tan⁡(34∘)⋅tan⁡(11∘)+tan⁡(34∘)+tan⁡(11∘)+1]\Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {\tan \left( {34^\circ } \right) \cdot \tan \left( {11^\circ } \right) + \tan \left( {34^\circ } \right) + \tan \left( {11^\circ } \right) + 1} \right]…………………………(2)\left( 2 \right)
Now as we know, tan⁡(45∘)=tan⁡(34∘+11∘)\tan \left( {45^\circ } \right) = \tan \left( {34^\circ + 11^\circ } \right), so using the identity, we get
1=tan⁡34∘+tan⁡11∘1−tan⁡34∘⋅tan⁡11∘1 = \dfrac{{\tan 34^\circ + \tan 11^\circ }}{{1 - \tan 34^\circ \cdot \tan 11^\circ }}
⇒1=tan⁡34∘+tan⁡11∘1−tan⁡34∘⋅tan⁡11∘\Rightarrow 1 = \dfrac{{\tan 34^\circ + \tan 11^\circ }}{{1 - \tan 34^\circ \cdot \tan 11^\circ }}
⇒1−tan⁡34∘⋅tan⁡11∘=tan⁡34∘+tan⁡11∘\Rightarrow 1 - \tan 34^\circ \cdot \tan 11^\circ = \tan 34^\circ + \tan 11^\circ
Now substituting the value of tan⁡34∘+tan⁡11∘\tan 34^\circ + \tan 11^\circ in equation number (2)\left( 2 \right), we get
⇒f(11∘)⋅f(34∘)=14[tan⁡(34∘)⋅tan⁡(11∘)+1−tan⁡(34∘)⋅tan⁡(11∘)+1]\Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {\tan \left( {34^\circ } \right) \cdot \tan \left( {11^\circ } \right) + 1 - \tan \left( {34^\circ } \right) \cdot \tan \left( {11^\circ } \right) + 1} \right]
Now eliminating the same constants, we get,
⇒f(11∘)⋅f(34∘)=14[1+1]\Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{4}\left[ {1 + 1} \right]
⇒f(11∘)⋅f(34∘)=24\Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{2}{4}
Dividing numerator and denominator by 22, we get
⇒f(11∘)⋅f(34∘)=12\Rightarrow f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) = \dfrac{1}{2}
Hence, the required value of f(11∘)⋅f(34∘)f\left( {11^\circ } \right) \cdot f\left( {34^\circ } \right) is 12\dfrac{1}{2}.

Hence, option A is the correct option.

Note:
If we are aware of all the basic and trigonometric formulas, then this question can be solved easily. This question can be a bit tricky at the beginning because it’s quite tough to recognize that from where we should begin solving this question. Also, keeping in mind the fact that the numerator can be rewritten using the formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab and other basic formulas. We should be aware of calculation mistakes in these types of questions as it is possible that we would use ‘plus’ sign instead of using a ‘minus’ and vice-versa in various formulas. Also, the fact that tan⁡(45∘)=tan⁡(34∘+11∘)\tan \left( {45^\circ } \right) = \tan \left( {34^\circ + 11^\circ } \right) helped us to further solve this question. If this basic addition would have not striked the mind then, this question could be a tough one to solve. Hence, all these facts make this question quite necessary.